使用PHP和HTML在每个表行上生成的按钮
[英]Generated button on each table row with PHP and HTML
问题描述
I have this code to generate buttons on each table row with PHP and HTML: 
我有以下代码使用PHP和HTML在每个表行上生成按钮：
<!DOCTYPE HTML>
<h2 class="text-center">Consulta</h2>
<br>
<table class="table table-striped">
<tr class="info">
<td class="black"><span style="font-weight:bold">#</span></td>
<td class="black"><span style="font-weight:bold">Data</span></td>
<td class="black"><span style="font-weight:bold">Condomínio</span></td>
<td class="black"><span style="font-weight:bold">Entrada</span></td>
<td class="black"><span style="font-weight:bold">Título</span></td>
<td class="black"><span style="font-weight:bold">Autor</span></td>
<td class="black"><span style="font-weight:bold">Estado</span></td>
<td class="black"><span style="font-weight:bold"></span></td>
</tr>
<?php
$select = "SELECT * FROM database";
$get = mysqli_query($cnn, $select) or die ('error');
while($row = mysqli_fetch_array($get, MYSQLI_ASSOC)) {
echo "<tr>";
echo "<td>";
echo $row['id'];
echo "</td><td>";
echo $row['data'];
echo "</td><td>";
echo $row['condominio'];
echo "</td><td>";
echo $row['entrada'];
echo "</td><td>";
echo $row['titulo'];
echo "</td><td>";
echo $row['autor'];
echo "</td><td>";
echo $row['estado'];
echo "</td><td>";
$html = '<form role="form" class="form-inline" method="POST" enctype="multipart/form-data" action="thispage.php">';
echo $html;
echo '<button type="submit" name="alter'.$row['id'].'" id="alter'.$row['id'].'" class="btn btn-default btn-xl">alter'.$row['id'].'</a></button>';
echo '</form>';
echo "</td>";
echo "</tr>";
}
echo "</table>";
?>
</html>
And, of course, I want to use the buttons, but the code below won't work for me: 而且，当然，我想使用按钮，但是下面的代码对我不起作用：
<?php
$var = "row['id']";
$var2 = "alter".$var;
if (isset($_POST['$var2'])) {
echo('some code');
}
?>
I saw some solutions with Javascript but I can only use PHP and HTML, so if anyone could help I would appreciate it. 我看到了一些使用Javascript的解决方案，但我只能使用PHP和HTML，因此，如果有人可以帮助，我将不胜感激。
3 个解决方案
解决方案1
1 2016-02-24 11:51:28
in simple way replace button to link 以简单的方式替换按钮链接
when you get the data put a link like that 当您获取数据时，放置一个类似的链接
echo " <a href='?id=".$row['id']."'>'alter'".$row['id']."</a>"
in next page use it like that 在下一页像这样使用它
if(isset($_GET['id']))
{
do something
}
解决方案2
0 2016-02-24 11:22:21
if (isset($_POST[$var2])) {
echo('some code');
}
Try this. 尝试这个。 I guess $var2 is treated as a string since it is covered with single quote 我猜$var2被当作一个字符串，因为它$var2引号覆盖
解决方案3
0 2016-02-24 12:25:11
If $var2 should read "alter1", "alter2", "alter3" etc. the line, 如果$ var2应该将“ alter1”，“ alter2”，“ alter3”等读为该行，
$var = "row['id']";
should read, 应该读，
$var = $row['id'];
Craig 克雷格
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